The electrostatic potential inside a charged& | vedclass

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Question

Electric filed, $E=-\frac{d \phi}{d r}=-2 a r$

By Gauss's law, $E .4 \pi r^{2}=\frac{q_{i n}}{\epsilon_{0}}$

$\Rightarrow q_{i n}=(-2 a r) 4

Vedclass · Accepted Answer

$6\,a{\varepsilon _0}$

Vedclass · Answer

Electric filed, $E=-\frac{d \phi}{d r}=-2 a r$

By Gauss's law, $E .4 \pi r^{2}=\frac{q_{i n}}{\epsilon_{0}}$

$\Rightarrow q_{i n}=(-2 a r) 4 \pi r^{2} \epsilon_{0}=-8 \pi \epsilon_{0} a r^{3}$

Now $\frac{d q_{i n}}{d r}=-24 \pi \epsilon_{0} a r^{2}$ and $V=\frac{4}{3} \pi r^{3}, \frac{d V}{d r}=4 \pi r^{2}$

Charge density, $ho=\frac{d q_{i n}}{d V}=\frac{d q_{i n}}{d r} 	imes \frac{d r}{d V}=\left(-24 \pi \epsilon_{0} a r^{2}ight) 	imes \frac{1}{4 \pi r^{2}}=-6 \epsilon_{0} a$

The electrostatic potential inside a charged spherical ball is given by : $V = b -ar^2$, where $r$ is the distance from the centre ; $a$ and $b$ are constants. Then, the charge density inside the ball is :

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